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Greeley [361]
2 years ago
12

A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta

l. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t = 1.03 s? Repeat for the (c) horizontal and (d) vertical components at t = 1.71 s, and for the (e) horizontal and (f) vertical components at t = 5.44 s. Assume that the catapult is positioned on a plain horizontal ground.
Physics
1 answer:
larisa [96]2 years ago
6 0

Answer:

Part a)

x = 15.76 m

Part b)

y = 7.94 m

Part c)

x = 26.16 m

Part d)

y = 7.49 m

Part e)

x = 83.23 m

Part f)

y = -75.6 m

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

v_x = 19.9 cos39.9

v_x = 15.3 m/s

similarly we have

v_y = 19.9 sin39.9

v_y = 12.76 m/s

Part a)

Horizontal displacement in 1.03 s

x = v_x t

x = (15.3)(1.03)

x = 15.76 m

Part b)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.03) - 4.9(1.03)^2

y = 7.94 m

Part c)

Horizontal displacement in 1.71 s

x = v_x t

x = (15.3)(1.71)

x = 26.16 m

Part d)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.71) - 4.9(1.71)^2

y = 7.49 m

Part e)

Horizontal displacement in 5.44 s

x = v_x t

x = (15.3)(5.44)

x = 83.23 m

Part f)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(5.44) - 4.9(5.44)^2

y = -75.6 m

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Varvara68 [4.7K]

The resultant displacement of the man is 109.77 km in the direction N60°E.

<h3>Displacement</h3>

Displacement is the distance travelled in a specified direction.

To calculate displacement, the straight line from starting point to end point of travel is taken and calculated.

<h3>Resultant displacement of the man </h3>

In the example above, a man walks 95 km, East, then 55 km, north.

The two distances form a right-angled triangle with two sides 95 and 55 units. The hypotenuse gives the resultant displacement, D.

Using Pythagoras rule:

D^2 = 95^2 + 55^2

D^2 = 12050

D = 109.77

Thus, the resultant displacement is 109.77 km

To calculate the direction:

Let the direction be y

y + x = 90°

tan x = 55/95

tanx x = 0.578

x = 30°

Then, y = 90 - 30

y = 60°

Therefore, the resultant displacement of the man is 109.77 km in the direction N60°E.

Learn more about displacement at: brainly.com/question/321442

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2 years ago
6. 498.82 mg comverted to kg​
Ivanshal [37]

Answer:

0.00049882 kg

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6 0
2 years ago
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It is difficult to walk on a smooth floor why​
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A U-shaped tube, open to the air on both ends, contains mercury. Water is poured into the left arm until the water column is 10.
Masja [62]

Answer:

0.368 cm

Explanation:

x = distance by which the mercury rise

d = depth of the water = 10 cm = 0.10 m

ρ = density of water = 1000 kgm⁻³

ρ' = density of mercury = 13600 kgm⁻³

P₀ = atmospheric pressure

Using equilibrium of pressure on both side

P₀ + ρ g d = P₀ + ρ' g (2x)

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8 0
2 years ago
1) A plane's velocity increases from 40 m/s to 100 m/s over a 10 second interval. What is the plane's average acceleration for t
Yakvenalex [24]

Answer:

average acceleration = 6 \frac{m}{s^2}

Explanation:

Recall that the average acceleration (a)  is defined by the change in velocity from an initial velocity (v_i), to a final velocity (v_f) over the time (t) it took that change to happen. Then, in mathematical terms this is:

a=\frac{v_f-v_i}{t}

with our information this becomes:

a=\frac{v_f-v_i}{t} = \frac{100-40}{10}=6\,\frac{m}{s^2}

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